FRACTIONAL ELLIPTIC EQUATIONS WITH SIGN-CHANGING …Key words and phrases. Non-local operator; singular nonlinearity; Nehari manifold; sign-changing weight function. c 2016 Texas State

Electronic Journal of Differential Equations, Vol. 2016 (2016), No. 145, pp. 1–23.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

FRACTIONAL ELLIPTIC EQUATIONS WITH SIGN-CHANGINGAND SINGULAR NONLINEARITY

SARIKA GOYAL, KONIJETI SREENADH

Abstract. In this article, we study the fractional Laplacian equation withsingular nonlinearity

(−∆)su = a(x)u−q + λb(x)up in Ω,

u > 0 in Ω, u = 0 in ∂Ω,

where Ω is a bounded domain in Rn with smooth boundary ∂Ω, n > 2s, s ∈(0, 1), λ > 0. Using variational methods, we show existence and multiplicity

of positive solutions.

1. Introduction

Let Ω ⊂ Rn be a bounded domain with smooth boundary, n > 2s and s ∈ (0, 1).We consider the fractional elliptic problem with singular nonlinearity

(−∆)su = a(x)u−q + λb(x)up in Ω,u > 0 in Ω, u = 0 in ∂Ω.

(1.1)

We use the following assumptions on a and b:

(A1) a : Ω ⊂ Rn → R such that 0 < a ∈ L2∗s

2∗s−1+q (Ω).(A2) b : Ω ⊂ Rn → R is a sign-changing function such that b+ 6≡ 0 and b(x) ∈

L2∗s

2∗s−1−p (Ω).Here λ > 0 is a parameter, 0 < q < 1 < p < 2∗s − 1, with 2∗s = 2n

n−2s , known asfractional critical Sobolev exponent and where (−∆)s is the fractional Laplacianoperator in Ω with zero Dirichlet boundary values on ∂Ω.

To define the fractional Laplacian operator (−∆)s in Ω, let λk, φk be theeigenvalues and the corresponding eigenfunctions of −∆ in Ω with zero Dirichletboundary values on ∂Ω

(−∆)sφk = λkφk in Ω, φk = 0 on ∂Ω.

normalized by ‖φk‖L2(Ω) = 1. Then one can define the fractional Laplacian (−∆)s

for s ∈ (0, 1) by

(−∆)su =∞∑k=1

λskckφk,

2010 Mathematics Subject Classification. 35A15, 35J75, 35R11.Key words and phrases. Non-local operator; singular nonlinearity; Nehari manifold;

sign-changing weight function.c©2016 Texas State University.

Submitted January, 21, 2016. Published June 14, 2016.

1

2 S. GOYAL, K. SREENADH EJDE-2016/145

which clearly maps

Hs0(Ω) :=

u =

∞∑k=1

ckφk ∈ L2(Ω) : ‖u‖Hs0 (Ω) =( ∞∑k=1

λskc2k

)1/2

<∞

into L2(Ω). Moreover Hs0(Ω) is a Hilbert space endowed with an inner product⟨ ∞∑

k=1

ckφk,

∞∑k=1

dkφk⟩Hs0 (Ω)

=∞∑k=1

λskckdkφk, if∞∑k=1

ckφk,

∞∑k=1

dkφk ∈ Hs0(Ω).

Definition 1.1. A function u ∈ Hs0(Ω) such that u(x) > 0 in Ω is a solution of

(1.1) such that for every function v ∈ Hs0(Ω), it holds∫

Ω

(−∆)s/2u(−∆)s/2vdx =∫

Ω

a(x)u−qvdx+ λ

∫Ω

b(x)upvdx.

Associated with (1.1), we consider the energy functional for u ∈ Hs0(Ω), u > 0

in Ω such that

Iλ(u) =∫

Ω

|(−∆)s/2u|2dx− 11− q

∫Ω

a(x)|u|1−qdx− λ

p+ 1

∫Ω

b(x)|u|p+1dx.

The fractional power of Laplacian is the infinitesimal generator of Levy stablediffusion process and arise in anomalous diffusions in plasma, population dynamics,geophysical fluid dynamics, flames propagation, chemical reactions in liquids andAmerican options in finance. For more details, we refer to [3, 14] and referencetherein.

Recently the study of existence, multiplicity of solutions for fractional ellipticequations attracted a lot of interest by many researchers. Among the works dealingwith fractional elliptic equations we cite [6, 9, 21, 22, 23, 24, 25, 26] and referencestherein, with no attempt to provide a complete list. Caffarelli and Silvestre [8]gave a new formulation of fractional Laplacian through Dirichlet-Neumann maps.This formulation transforms problems involving the fractional Laplacian into a localproblem which allows one to use the variational methods.

On the other hand, there are some works where multiplicity results are shownusing the structure of associated Nehari manifold. In [15, 16] authors studiedsubcritical problems and in [28] the authors obtained the existence of multiplicityfor critical growth nonlinearity. In the case of the square root of Laplacian, themultiplicity results for sublinear and superlinear type of nonlinearity with sign-changing weight functions are studied in [7, 27].

In the local setting, s = 1, the paper by Crandall, Robinowitz and Tartar [12]is the starting point on semilinear problem with singular nonlinearity. There is alarge body of literature on singular problems, see [1, 2, 11, 12, 17, 18, 19, 20] andreference therein. Recently, Chen and Chen in [10] studied the following problemwith singular nonlinearity

−∆u− λ

|x|2= h(x)u−q + µW (x)up in Ω \ 0, u > 0 in Ω \ 0, u = 0 on ∂Ω,

where 0 ∈ Ω ⊂ Rn(n ≥ 3) is a bounded smooth domain with smooth boundary,0 < λ < (n−2)2

4 and 0 < q < 1 < p < n+2n−2 . Also h, W both are continuous

functions in Ω with h > 0 and W is sign-changing. By variational methods, theyshowed that there exists Tλ such that for µ ∈ (0, Tλ) the above problem has twopositive solutions.

EJDE-2016/145 FRACTIONAL LAPLACIAN WITH SINGULAR NONLINEARITY 3

In case of the fractional Laplacian, Fang [13] proved the existence of a solutionof the singular problem

(−∆)sw = w−p, u > 0 in Ω, u = 0 in Rn \ Ω,

with 0 < p < 1, using the method of sub and super solution. Recently, Barrios,Peral and et al [4] extend the result of [13]. They studied the existence result forthe singular problem

(−∆)su = λf(x)uγ

+Mup, u > 0 in Ω, u = 0 in Rn \ Ω,

where Ω is a bounded smooth domain of Rn, n > 2s, 0 < s < 1, γ > 0, λ > 0,p > 1 and f ∈ Lm(Ω), m ≥ 1 is a nonnegative function. For M = 0, they provedthe existence of solution for every γ > 0 and λ > 0. For M = 1 and f ≡ 1, theyshowed that there exist Λ such that it has a solution for every 0 < λ < Λ, andhave no solution for λ > Λ. Here the authors first studied the uniform estimates ofsolutions un of the regularized problems

(−∆)su = λf(x)

(u+ 1n )γ

+ up, u > 0 in Ω, u = 0 in Rn \ Ω. (1.2)

Then they obtained the solutions by taking limit in the regularized problem (1.2).As far as we know, there is no work related to fractional Laplacian for singular

nonlinearity and sign-changing weight functions. So, in this paper, we study themultiplicity results for problem (1.1) for 0 < q < 1 < p < 2∗s − 1 and λ > 0. Thiswork is motivated by the work of Chen and Chen in [10]. Due to the singularityof problem, it is not easy to deal the problem (1.1) as the associated functional isnot differentiable even in sense of Gateaux and the strong maximum principle isnot applicable to show the positivity of solutions. Moreover one can not directlyextend all the results from Laplacian case to fractional Laplacian, due to the non-local behavior of the operator and the bounded support of the test function is notpreserved. To overcome these difficulties, we first use the Cafferelli and Silvestre[9] approach to convert the problem (1.1) into the local problem. Then we use thevariational technique to study the local problem as in [10]. In this paper, the proofsof some Lemmas follow the similar lines as in [10] but for completeness, we give thedetails.

The article is organized as follows: In section 2 we present some preliminaries onextension problem and necessary weighted trace inequalities required for variationalsettings. We also state our main results. In section 3, we study the decompositionof Nehari manifold and local charts using the fibering maps. In Section 4, we showthe existence of a nontrivial solutions and show how these solutions arise out ofnature of Nehari manifold.

We will use the following notation throughout this paper: The same symbol

‖ · ‖ denotes the norms in the three spaces L2∗s

2∗s−1−q (Ω), L2∗s

2∗s−1−p (Ω), and Hs0,L(CΩ)

defined by (2.1). Also S := ksS(s, n) where S(s, n) is the best constant of Sobolevembedding (see (2.2)).

2. Preliminaries and main results

In this section we give some definitions and functional settings. At the end ofthis section, we state our main results. To state our main result, we introduce some


notation and basic preliminaries results. Denote the upper half-space in Rn+1 by

Rn+1+ := z = (x, y) = (x1, x2, . . . , xn, y) ∈ Rn+1|y > 0,

the half cylinder standing on a bounded smooth domain Ω ⊂ Rn by CΩ := Ω ×(0,∞) ⊂ Rn+1

+ and its lateral boundary is denoted by ∂LCΩ = ∂Ω× [0,∞). Definethe function space Hs

0,L(CΩ) as the completion of C∞0,L(CΩ) = w ∈ C∞(CΩ) : w =0 on ∂LCΩ under the norm

‖w‖Hs0,L(CΩ) =(ks

∫CΩy1−2s|∇w(x, y)|2 dx dy

)1/2

, (2.1)

where ks := 21−2sΓ(1−s)Γ(s) is a normalization constant. Then it is a Hilbert space

endowed with the inner product

〈w, v〉Hs0,L(CΩ) = ks

∫Ω×0

y1−2s∇w∇v dx dy.

If Ω is a smooth bounded domain then it is verified that (see [8, Proposition 2.1],[6, Proposition 2.1], [26, section 2])

Hs0(Ω) := u = tr |Ω×0w : w ∈ Hs

0,L(CΩ)

and there exists a constant C > 0 such that

‖w(·, 0)‖Hs0 (Ω) ≤ C‖w‖Hs0,L(CΩ) for all w ∈ Hs0,L(CΩ).

Now we define the extension operator and fractional Laplacian for functions inHs

0(Ω).

Definition 2.1. Given a function u ∈ Hs0(Ω), we define its s-harmonic extension

w = Es(u) to the cylinder CΩ as a solution of the problem

div(y1−2s∇w) = 0 in CΩ,w = 0 on ∂LCΩ,w = u on Ω× 0.

Definition 2.2. For any regular function u ∈ Hs0(Ω), the fractional Laplacian

(−∆)s acting on u is defined by

(−∆)su(x) = −ks limy→0+

y1−2s ∂w

∂y(x, y) for all (x, y) ∈ CΩ.

From [5] and [9], the map Es(·) is an isometry between Hs0(Ω) and Hs

0,L(CΩ).Furthermore, we have

(1)‖(−∆)su‖H−s(Ω) = ‖u‖Hs0 (Ω) = ‖Es(u)‖Hs0,L(CΩ),

where H−s(Ω) denotes the dual space of Hs0(Ω);

(2) For any w ∈ Hs0,L(CΩ), there exists a constant C independent of w such

that‖ trΩ w‖Lr(Ω) ≤ C‖w‖Hs0,L(CΩ)

holds for every r ∈ [2, 2nn−2s ]. Moreover, Hs

0,L(CΩ) is compactly embeddedinto Lr(Ω) for r ∈ [2, 2n

n−2s ).


Lemma 2.3. For every 1 ≤ r ≤ 2nn−2s and every w ∈ Hs

0,L(CΩ), it holds(∫Ω×0

|w(x, 0)|rdx)2/r

≤ Cks∫CΩy1−2s|∇w(x, y)|2 dx dy,

where the constant C depends on r, s, n and |Ω|.

Lemma 2.4. For every w ∈ Hs(Rn+1+ ), it holds

S(s, n)(∫

Rn|u(x)|

2nn−2s dx

)n−2sn ≤

∫Rn+1

+

y1−2s|∇w(x, y)|2 dx dy, (2.2)

where u = trΩ w. The constant S(s, n) is known as the best constant and takes thevalue

S(s, n) =2πsΓ( 2−2s

2 )Γ(n+2s2 )(Γ(n2 ))

2sn

Γ(s)Γ(n−2s2 )(Γ(n))

2sn

.

Now we can transform the nonlocal problem (1.1) into the local problem

−div(y1−2s∇w) = 0 in CΩ := Ω× (0,∞),

w = 0 on ∂LCΩ, w > 0 on Ω× 0,∂w

∂v2s= a(x)w−q + λb(x)wp on Ω× 0,

(2.3)

where ∂w∂v2s := −ks limy→0+ y1−2s ∂w

∂y (x, y), for all x ∈ Ω.

Definition 2.5. A weak solution of (2.3) is a function w ∈ Hs0,L(CΩ), w > 0 in

Ω× 0 such that for every v ∈ Hs0,L(CΩ),

ks

∫CΩy1−2s∇w∇v dx dy

=∫

Ω×0a(x)(w−qv)(x, 0)dx+ λ

∫Ω×0

b(x)(wpv)(x, 0)dx.

If w satisfies (2.3), then u = trΩ w = w(x, 0) ∈ Hs0(Ω) is a weak solution to problem

(1.1).

Let c = 1−2s, then the associated functional Jλ : Hs0,L(CΩ)→ R to the problem

(2.3) is

Jλ(w) =ks2

∫CΩyc|∇w|2 dx dy − 1

1− q

∫Ω×0

a(x)|w|1−qdx

− λ

p+ 1

∫Ω×0

b(x)|w(x, 0)|p+1dx.

Now for w ∈ Hs0,L(CΩ), we define the fiber map φw : R+ → R as

φw(t) = Jλ(tw) =t2

2‖w‖2 − 1

1− q

∫Ω×0

a(x)|tw|1−qdx

− λtp+1

p+ 1

∫Ω×0

b(x)|w(x, 0)|p+1dx.

Then

φ′w(t) = t‖w‖2 − t−q∫

Ω×0a(x)|w(x, 0)|1−qdx− λtp

∫Ω×0

b(x)|w(x, 0)|p+1dx,


φ′′w(t) = ‖w‖2 + qt−q−1

∫Ω×0

a(x)|w(x, 0)|1−qdx

− pλtp−1

∫Ω×0

b(x)|w(x, 0)|p+1dx.

It is easy to see that the energy functional Jλ is not bounded below on the spaceHs

0,L(CΩ). But we will show that it is bounded below on an appropriate subset ofHs

0,L(CΩ) and a minimizer on subsets of this set gives rise to solutions of (2.3). Inorder to obtain the existence results, we define

Nλ : = w ∈ Hs0,L(CΩ) : 〈J ′λ(w), w〉 = 0

=w ∈ Hs

0,L(CΩ) : ‖w‖2 =∫

Ω×0a(x)|w(x, 0)|1−qdx

+ λ

∫Ω×0

b(x)|w(x, 0)|p+1dx.

Note that w ∈ Nλ if w is a solution of (2.3). Also one can easily see that tw ∈ Nλif and only if φ′w(t) = 0 and in particular, w ∈ Nλ if and only if φ′w(1) = 0. Inorder to obtain our result, we decompose Nλ with N±λ , N 0

λ as follows:

N±λ = w ∈ Nλ : φ′′w(1) ≷ 0

=w ∈ Nλ : (1 + q)‖w‖2 ≷ λ(p+ q)

∫Ω×0

b(x)|w(x, 0)|p+1 dx,

N 0λ = w ∈ Nλ : φ′′w(1) = 0

=w ∈ Nλ : (1 + q)‖w‖2 = λ(p+ q)

∫Ω×0

b(x)|w(x, 0)|p+1 dx.

Inspired by [9] and [10], we show that how variational methods can be used toestablished some existence and multiplicity results. Our results are as follows:

Theorem 2.6. Suppose that λ ∈ (0,Λ), where

Λ :=1 + q

p+ q

(p− 1p+ q

) p−11+q 1‖b‖

( Sp+q

‖a‖p−1

)1/(1+q)

.

Then problem (2.3) has at least two solutions w0 ∈ N+λ , W0 ∈ N−λ with ‖W0‖ >

‖w0‖. Moreover, u0(x) = w0(·, 0) ∈ Hs0(Ω) and U0(x) = W0(·, 0) ∈ Hs

0(Ω) arepositive solutions of the problem (1.1).

Next, we obtain the blow up behavior of the solution Wε ∈ N−λ of problem (2.3)with p = 1 + ε as ε→ 0+.

Theorem 2.7. let Wε ∈ N−λ be the solution of problem (2.3) with p = 1 + ε, whereλ ∈ (0,Λ), then

‖Wε‖ > Cε(Λλ

)1/ε,

where

Cε =(

1 +1 + q

ε

)1/(1+q)

‖a‖1/(q+1)( 1√

S

) 1−q1+q →∞ as ε→ 0+.

Namely, Wε blows up faster than exponentially with respect to ε.


We remark that if w is a positive solution of the problem

(−∆)su = a(x)u−q + λb(x)up in Ωu > 0 in Ω, u = 0 on ∂Ω.

Then one can easily see that v = λ1/(p−1)u is a positive solution of the problem

(−∆)sv = λ1+qp−1 a(x)v−q + b(x)vp in Ω

v > 0 in Ω, v = 0 in ∂Ω.(2.4)

That is, the problem (2.4) has two positive solution for λ ∈ (0,Λp−1).

3. Fibering map analysis

In this section, we show that N±λ is nonempty and N 0λ = 0. Moreover, Jλ is

bounded below and coercive.

Lemma 3.1. Let λ ∈ (0,Λ). Then for each w ∈ Hs0,L(CΩ) with∫

Ω×0a(x)|w(x, 0)|1−qdx > 0,

we have the following:(i)∫

Ω×0 b(x)|w(x, 0)|p+1 dx ≤ 0, then there exists a unique t1 < tmax suchthat t1w ∈ N+

λ and Jλ(t1w) = inft>0 Jλ(tw);(ii)

∫Ω×0 b(x)|w(x, 0)|p+1 dx > 0, then there exists a unique t1 and t2 with

0 < t1 < tmax < t2 such that t1w ∈ N+λ , t2w ∈ N−λ and Jλ(t1w) =

inf0≤t≤tmax Jλ(tw), Jλ(t2w) = supt≥t1 Jλ(tw).

Proof. For t > 0, we define

ψw(t) = t1−p‖w‖2−t−p−q∫

Ω×0a(x)|w(x, 0)|1−q dx−λ

∫Ω×0

b(x)|w(x, 0)|p+1 dx.

One can easily see that ψw(t)→ −∞ as t→ 0+. Now

ψ′w(t) = (1− p)t−p‖w‖2 + (p+ q)t−p−q−1

∫Ω×0

a(x)|w(x, 0)|1−q dx.

ψ′′w(t) = −p(1− p)t−p−1‖w‖2

− (p+ q)(p+ q + 1)t−p−q−2

∫Ω×0

a(x)|w(x, 0)|1−q dx.

Then ψ′w(t) = 0 if and only if t = tmax := [ (p−1)‖w‖2(p+q)

RΩ×0 a(x)|w(x,0)|1−q dx ]−1/(1+q).

Also

ψ′′w(tmax) = p(p− 1)[ (p− 1)‖w‖2

(p+ q)∫

Ω×0 a(x)|w(x, 0)|1−q dx

] p+1q+1 ‖w‖2

− (p+ q)(p+ q + 1)[ (p− 1)‖w‖2

(p+ q)∫

Ω×0 a(x)|w(x, 0)|1−q dx

] p+q+2q+1

×∫

Ω×0a(x)|w(x, 0)|1−q dx

= −‖w‖2(p− 1)(1 + q)[ (p− 1)‖w‖2

(p+ q)∫

Ω×0 a(x)|w(x, 0)|1−q dx

] p+1q+1

< 0.


Thus ψw achieves its maximum at t = tmax. Now using the Holder’s inequality andSobolev inequality (2.2), we obtain∫

Ω×0a(x)|w(x, 0)|1−q dx

≤[ ∫

Ω×0|a(x)|

2∗s2∗s−1+q

] 2∗s+q−12∗s

[ ∫Ω×0

|w(x, 0)|2∗sdx] 1−q

2∗s

≤ ‖a‖(‖w‖√

S

)1−q.

(3.1)

∫Ω×0

b(x)|w(x, 0)|p+1 dx

≤[ ∫

Ω×0|b(x)|

2∗s2∗s−1−p

] 2∗s−p−12∗s

[ ∫Ω×0

|w(x, 0)|2∗sdx] p+1

2∗s

≤ ‖b‖(‖w‖√

S

)p+1

.

(3.2)

Using (3.1) and (3.2), we obtain

ψw(tmax)

=1 + q

p+ q

(p− 1p+ q

) p−11+q ‖w‖

2(p+q)1+q

[∫

Ω×0 a(x)|w(x, 0)|1−q dx]p−11+q

− λ∫

Ω×0b(x)|w(x, 0)|p+1 dx

≥[1 + q

p+ q

(p− 1p+ q

) p−11+q( (√S)(1−q)

‖a‖) p−1

1+q − λ‖b‖( 1√

S

)p+1]‖w‖p+1

≡ Eλ‖w‖p+1,

(3.3)where

Eλ :=[1 + q

p+ q

(p− 1p+ q

) p−11+q( (√S)(1−q)

‖a‖

) p−11+q − λ‖b‖

( 1√S

)p+1].

Then we see that Eλ = 0 if and only if λ = Λ, where

Λ :=1 + q

p+ q

(p− 1p+ q

) p−11+q 1‖b‖

( Sp+q

‖a‖p−1

)1/(1+q)

.

Thus for λ ∈ (0,Λ), we have Eλ > 0, and therefore it follows from (3.3) thatψw(tmax) > 0.

(i) If∫

Ω×0 b(x)|w(x, 0)|p+1 dx ≥ 0, then

ψw(t)→ −λ∫

Ω×0b(x)|w(x, 0)|p+1 dx < 0

as t → ∞. Consequently, ψw(t) has exactly two points 0 < t1 < tmax < t2 suchthat

ψw(t1) = 0 = ψw(t2) and ψ′w(t1) > 0 > ψ′w(t2).

From the definition of N±λ , we get t1w ∈ N+λ and t2w ∈ N−λ . Now φ′w(t) = tpψw(t).

Thus φ′w(t) < 0 in (0, t1), φ′w(t) > 0 in (t1, t2) and φ′w(t) < 0 in (t2,∞). HenceJλ(t1w) = inf0≤t≤tmax Jλ(tw), Jλ(t2w) = supt≥t1 Jλ(tw).


(ii) If∫

Ω×0 b(x)|w(x, 0)|p+1 dx < 0 and

ψw(t)→ −λ∫

Ω×0b(x)|w(x, 0)|p+1 dx > 0

as t→∞. Consequently, ψw(t) has exactly one point 0 < t1 < tmax such that

ψw(t1) = 0 and ψ′w(t1) > 0.

Now φ′w(t) = tpψw(t). Then φ′w(t) < 0 in (0, t1), φ′w(t) > 0 in (t1,∞). ThusJλ(t1w) = inft≥0 Jλ(tw). Hence, it follows that t1w ∈ N+

λ .

Corollary 3.2. Suppose that λ ∈ (0,Λ), then N±λ 6= ∅.

Proof. From (A1) and (A2), we can choose w ∈ Hs0,L(CΩ) \ 0 such that∫

Ω×0a(x)|w(x, 0)|1−q dx > 0 and

∫Ω×0

b(x)|w(x, 0)|p+1 dx > 0.

Then by (ii) of Lemma 3.1, there exists a unique t1 and t2 such that t1w ∈ N+λ ,

t2w ∈ N−λ . In conclusion, N±λ 6= ∅.

Lemma 3.3. For λ ∈ (0,Λ), we have N 0λ = 0.

Proof. We prove this by contradiction. Assume that there exists 0 6≡ w ∈ N 0λ .

Then it follows from w ∈ N 0λ that

(1 + q)‖w‖2 = λ(p+ q)∫

Ω×0b(x)|w(x, 0)|p+1 dx

and consequently

0 = ‖w‖2 −∫

Ω×0a(x)|w(x, 0)|1−q dx− λ

∫Ω×0

b(x)|w(x, 0)|p+1 dx

=p− 1p+ q

‖w‖2 −∫

Ω×0a(x)|w(x, 0)|1−q dx.

Therefore, as λ ∈ (0,Λ) and w 6≡ 0, we use similar arguments as those in (3.3) toobain

0 < Eλ‖w‖p+1

≤ 1 + q

p+ q

(p− 1p+ q

) p−11+q ‖w‖

2(p+q)1+q[ ∫

Ω×0 a(x)|w(x, 0)|1−q dx] p−1

1+q

− λ∫

Ω×0b(x)|w(x, 0)|p+1 dx

=1 + q

p+ q

(p− 1p+ q

) p−11+q ‖w‖

2(p+q)1+q(

p−1p+q ‖w‖2

) p−11+q

− 1 + q

p+ q‖w‖2 = 0,

a contradiction. Hence w ≡ 0. That is, N 0λ = 0.

We note that Λ is also related to a gap structure in Nλ:


Lemma 3.4. Suppose that λ ∈ (0,Λ), then there exist a gap structure in Nλ:

‖W‖ > Aλ > A0 > ‖w‖ for all w ∈ N+λ , W ∈ N

−λ ,

where

Aλ =[ 1 + q

λ(p+ q)‖b‖(√S)p+1

]1/(p−1)

and A0 =[p+ q

p− 1‖a‖( 1√

S

)1−q]1/(q+1)

.

Proof. If w ∈ N+λ ⊂ Nλ, then

0 < (1 + q)‖w‖2 − λ(p+ q)∫

Ω×0b(x)|w(x, 0)|p+1dx

= (1 + q)‖w‖2 − (p+ q)[‖w‖2 −

∫Ω×0

a(x)|w(x, 0)|1−qdx]

= (1− p)‖w‖2 + (p+ q)∫

Ω×0a(x)|w(x, 0)|1−qdx.

Hence it follows from (3.1) that

(p− 1)‖w‖2 < (p+ q)∫

Ω×0a(x)|w(x, 0)|1−qdx ≤ (p+ q)‖a‖

(‖w‖√S

)1−q

which yields

‖w‖ <[p+ q

p− 1‖a‖( 1√

S

)1−q]1/(q+1)

≡ A0.

If W ∈ N−λ , then it follows from (3.2) that

(1 + q)‖W‖2 < λ(p+ q)∫

Ω×0b(x)|W (x, 0)|p+1dx ≤ λ(p+ q)‖b‖

(‖W‖√S

)p+1

which yields

‖W‖ >[ 1 + q

λ(p+ q)‖b‖(√S)p+1

]1/(p−1)

≡ Aλ.

Now we show that Aλ = A0 if and only if λ = Λ.

λ = Λ =1 + q

p+ q

(p− 1p+ q

) p−11+q 1‖b‖

( Sp+q

‖a‖p−1

)1/(1+q)

if and only if

Aλ = λ−1/(p−1)(1 + q

p+ q

)1/(p−1)( 1‖b‖

)1/(p−1)(√S) p+1p−1

=(p+ q

p− 1

)1/(1+q)

‖a‖1/(q+1)(√

S)− 2(p+q)

(1+q)(p−1) + p+1p−1

=[ (p+ q)‖a‖

(p− 1)(√S)1−q

]1/(q+1)

≡ A0.

Thus for all λ ∈ (0,Λ), we can conclude that

‖W‖ > Aλ > A0 > ‖w‖ for all w ∈ N+λ ,W ∈ N

−λ .

This completes the proof.

Lemma 3.5. Suppose that λ ∈ (0,Λ), then N−λ is a closed set in Hs0,L(CΩ)-topology.


Proof. Let Wk be a sequence in N−λ with Wk →W0 in Hs0,L(CΩ). Then we have

‖W0‖2 = limk→∞

‖Wk‖2

= limk→∞

[ ∫Ω×0

a(x)|Wk(x, 0)|1−qdx+ λ

∫Ω×0

b(x)|Wk(x, 0)|p+1dx]

=∫

Ω×0a(x)|W0(x, 0)|1−qdx+ λ

∫Ω×0

b(x)|W0(x, 0)|p+1dx

and

(1 + q)‖W0‖2 − λ(p+ q)∫

Ω×0b(x)|W0(x, 0)|p+1dx

= limk→∞

[(1 + q)‖Wk‖2 − λ(p+ q)

∫Ω×0

b(x)|Wk(x, 0)|p+1dx]≤ 0.

That is, W0 ∈ N−λ ∩N 0λ . Since Wk ⊂ N−λ , from Lemma 3.4 we have

‖W0‖ = limk→∞

‖Wk‖ ≥ A0 > 0,

which imply, W0 6= 0. It follows from Lemma 3.1, that W0 6∈ N 0λ for any λ ∈

(0,Λ). Thus W0 ∈ N−λ . Hence, N−λ is a closed set in Hs0,L(CΩ)-topology for any

λ ∈ (0,Λ).

Lemma 3.6. Let w ∈ N±λ . Then for any φ ∈ C∞0,L(CΩ), there exists a numberε > 0 and a continuous function f : Bε(0) := v ∈ Hs

0,L(CΩ) : ‖v‖ < ε → R+ suchthat

f(v) > 0, f(0) = 1, f(v)(w + vφ) ∈ N±λ for all v ∈ Bε(0).

Proof. We give the proof only for the case w ∈ N+λ , the case N−λ may be preceded

exactly. For any φ ∈ C∞0,L(CΩ), we define F : Hs0,L(CΩ)× R+ → R as follows:

F (v, r) = r1+q‖w + vφ‖2 −∫

Ω×0a(x)|(w + vφ)(x, 0)|1−q

− λrp+q∫

Ω×0b(x)|(w + vφ)(x, 0)|p+1.

Since w ∈ N+λ (⊂ Nλ), we have

F (0, 1) = ‖w‖2 −∫

Ω×0a(x)|w(x, 0)|1−qdx− λ

∫Ω×0

b(x)|w(x, 0)|p+1dx = 0,

and∂F

∂r(0, 1) = (1 + q)‖w‖2 − λ(p+ q)

∫Ω×0

b(x)|w(x, 0)|p+1dx > 0.

Applying the implicit function theorem at (0, 1), we have that there exists ε > 0such that for ‖v‖ < ε, v ∈ Hs

0,L(CΩ), the equation F (v, r) = 0 has a uniquecontinuous solution r = f(v) > 0. It follows from F (0, 1) = 0 that f(0) = 1 andfrom F (v, f(v)) = 0 for ‖v‖ < ε, v ∈ Hs

0,L(CΩ) that

0 = f1+q(v)‖w + vφ‖2 −∫

Ω×0a(x)|(w + vφ)(x, 0)|1−q

− λfp+q(v)∫

Ω×0b(x)|(w + vφ)(x, 0)|p+1


=(‖f(v)(w + vφ)‖2 −

∫Ω×0

a(x)|f(v)(w + vφ)(x, 0)|1−q

− λ∫

Ω×0b(x)|f(v)(w + vφ)(x, 0)|p+1

)/f1−q(v);

that is,f(v)(w + vφ) ∈ Nλ for all v ∈ Hs

0,L(CΩ), ‖v‖ < ε.

Since ∂F∂r (0, 1) > 0 and

∂F

∂r(v, f(v))

= (1 + q)fq(v)‖w + vφ‖2 − λ(p+ q)fp+q−1(v)∫

Ω×0b(x)|(w + vφ)(x, 0)|p+1dx

=(1 + q)‖f(v)(w + vφ)‖2 − λ(p+ q)

∫Ω×0 b(x)|f(v)(w + vφ)(x, 0)|p+1dx

f2−q(v),

we can take ε > 0 possibly smaller (ε < ε) such that for any v ∈ Hs0,L(CΩ), ‖v‖ < ε,

(1 + q)‖f(v)(w + vφ)‖2 − λ(p+ q)∫

Ω×0b(x)|f(v)(w + vφ)(x, 0)|p+1dx > 0;

that is,f(v)(w + vφ) ∈ N+

λ for all v ∈ Bε(0).This completes the proof.

Lemma 3.7. Jλ is bounded below and coercive on Nλ.

Proof. For w ∈ Nλ, from (3.1), we obtain

Jλ(w) =(1

2− 1p+ 1

)‖w‖2 −

( 11− q

− 1p+ 1

)∫Ω×0

a(x)|w(x, 0)|1−q dx

≥(1

2− 1p+ 1

)‖w‖2 −

( 11− q

− 1p+ 1

)‖a‖(‖w‖√

S

)1−q.

(3.4)

Now consider the function ρ : R+ → R as ρ(t) = αt2 − βt1−q, where α, β are bothpositive constants. One can easily show that ρ is convex(ρ′′(t) > 0 for all t > 0)with ρ(t) → 0 as t → 0 and ρ(t) → ∞ as t → ∞. ρ achieves its minimum attmin = [β(1−q)

2α ]1/(1+q) and

ρ(tmin) = α[β(1− q)

2α] 2

1+q − β[β(1− q)

2α] 1−q

1+q = −1 + q

2β

21+q

(1− q2α

) 1−q1+q

.

Applying ρ(t) with α =(

12 −

1p+1

), β =

(1

1−q −1p+1

)‖a‖(

1√S

)1−q and t = ‖w‖,w ∈ Nλ, we obtain from (3.4) that

lim‖w‖→∞

Jλ(w) ≥ limt→∞

ρ(t) =∞.

Thus Jλ is coercive on Nλ. Moreover, it follows from (3.4) that

Jλ(w) ≥ ρ(t) ≥ ρ(tmin) (a constant), (3.5)

i.e.,

Jλ(w) ≥ −1 + q

2β

21+q

(1− q2α

) 1−q1+q

= − 1 + q

(1− q)(p+ 1)

( (p+ q)‖a‖2(√S)1−q

) 21+q( 1p− 1

) 1−q1+q

.


Thus Jλ is bounded below on Nλ.

4. Existence of solutions in N±λNow from Lemma 3.5, N+

λ ∪N 0λ and N−λ are two closed sets in Hs

0,L(CΩ) providedλ ∈ (0,Λ). Consequently, the Ekeland variational principle can be applied to theproblem of finding the infimum of Jλ on both N+

λ ∪ N 0λ and N−λ . First, consider

wk ⊂ N+λ ∪N 0

λ with the following properties:

Jλ(wk) < infw∈N+

λ ∪N0λ

Jλ(w) +1k

(4.1)

Jλ(w) ≥ Jλ(wk)− 1k‖w − wk‖, ∀w ∈ N+

λ ∪N0λ . (4.2)

From Jλ(|w|) = Jλ(w), we may assume that wk ≥ 0 on CΩ.

Lemma 4.1. Show that the sequence wk is bounded in Nλ. Moreover, thereexists 0 6= w0 ∈ Hs

0,L(CΩ) such that wk w0 weakly in Hs0,L(CΩ).

Proof. By equations (3.5) and (4.1), we have

at2 − bt1−q = ρ(t) ≤ Jλ(w) < infw∈N+

λ ∪N0λ

Jλ(w) +1k≤ C5,

for sufficiently large k and a suitable positive constant. Hence putting t = wk inthe above equation, we obtain wk is bounded.

Let wk is bounded in Hs0,L(CΩ). Then, there exists a subsequence of wkk,

still denoted by wkk and w0 ∈ Hs0,L(CΩ) such that wk w0 weakly in Hs

0,L(CΩ),wk(·, 0)→ w0(·, 0) strongly in Lp(Ω) for 1 ≤ p < 2∗s and wk(·, 0)→ w0(·, 0) a.e. inΩ.

For any w ∈ N+λ , we have from 0 < q < 1 < p that

Jλ(w) =(1

2− 1

1− q

)‖w‖2 +

( 11− q

− 1p+ 1

)λ

∫Ω×0

b(x)|w(x, 0)|p+1 dx

<(1

2− 1

1− q

)‖w‖2 +

( 11− q

− 1p+ 1

)1 + q

p+ q‖w‖2

=( 1p+ 1

− 12

)1 + q

1− q‖w‖2 < 0,

which means that infN+λJλ < 0. Now for λ ∈ (0,Λ), we know from Lemma 3.1,

that N 0λ = 0. Together, these imply that wk ∈ N+

λ for k large and

infw∈N+

λ ∪N0λ

Jλ(w) ≤ infw∈N+

λ

Jλ(w) < 0.

Therefore, by weak lower semi-continuity of norm,

Jλ(w0) ≤ lim infk→∞

Jλ(wk) = infN+λ ∪N

0λ

Jλ < 0,

that is, w0 ≥ 0, w0 6≡ 0.

Lemma 4.2. Suppose wk ∈ N+λ such that wk w0 weakly in Hs

0,L(CΩ). Then forλ ∈ (0,Λ),

(1 + q)∫

Ω×0a(x)w1−q

0 (x, 0)dx− λ(p− 1)∫

Ω×0b(x)wp+1

0 (x, 0)dx > 0. (4.3)


Moreover, there exists a constant C2 > 0 such that

(1 + q)‖wk‖2 − λ(p+ q)∫

Ω×0b(x)wp+1

k (x, 0) dx ≥ C2 > 0. (4.4)

Proof. For wk ⊂ N+λ (⊂ Nλ), since

(1 + q)∫

Ω×0a(x)w1−q

0 (x, 0)dx− λ(p− 1)∫

Ω×0b(x)wp+1

0 (x, 0)dx

= limk→∞

[(1 + q)

∫Ω×0

a(x)w1−qk (x, 0) dx− λ(p− 1)

∫Ω×0

b(x)wp+1k (x, 0) dx

]= limk→∞

[(1 + q)‖wk‖2 − λ(p+ q)

∫Ω×0

b(x)wp+1k (x, 0) dx

]≥ 0,

we can argue by a contradiction and assume that

(1 + q)∫

Ω×0a(x)w1−q

0 (x, 0)dx− λ(p− 1)∫

Ω×0b(x)wp+1

0 (x, 0)dx = 0. (4.5)

Since wk ∈ Nλ, from the weak lower semi continuity of norm and (4.5) we have

0 = limk→∞

[‖wk‖2 −

∫Ω×0

a(x)w1−qk (x, 0) dx− λ

∫Ω×0

b(x)wp+1k (x, 0) dx

]≥ ‖w0‖2 −

∫Ω×0

a(x)w1−q0 (x, 0)dx− λ

∫Ω×0

b(x)wp+10 (x, 0)dx

=

‖w0‖2 − λp+q1+q

∫Ω×0 b(x)wp+1

0 (x, 0)dx

‖w0‖2 − p+qp−1

∫Ω×0 a(x)w1−q

0 (x, 0)dx.

Thus for any λ ∈ (0,Λ) and w0 6≡ 0, by similar arguments as those in (3.3) we havethat

0 < Eλ‖w0‖p+1

≤ 1 + q

p+ q

(p− 1p+ q

) p−11+q ‖w0‖

2(p+q)1+q[ ∫

Ω×0 a(x)w1−q0 (x, 0)dx

] p−11+q

− λ∫

Ω×0b(x)wp+1

0 (x, 0)dx

=1 + q

p+ q

(p− 1p+ q

) p−11+q ‖w0‖

2(p+q)1+q(

p−1p+q ‖w0‖2

) p−11+q

− 1 + q

p+ q‖w0‖2 = 0,

which is clearly impossible. Now by (4.3), we have that

(1 + q)∫

Ω×0a(x)w1−q

k (x, 0) dx− λ(p− 1)∫

Ω×0b(x)wp+1

k (x, 0) dx ≥ C2

for sufficiently large k and a suitable positive constant C2. This, together with thefact that wk ∈ Nλ we obtain equation (4.4).

Fix φ ∈ C∞0,L(CΩ) with φ ≥ 0. Then we apply Lemma 3.6 with w = wk ∈ N+λ

(k large enough such that (1−q)C1k < C2), we obtain a sequence of functions fk :

Bεk(0)→ R such that fk(0) = 1 and fk(w)(wk + wφ) ∈ N+λ for all w ∈ Bεk(0). It

follows from wk ∈ Nλ and fk(w)(wk + wφ) ∈ Nλ that

‖wk‖2 −∫

Ω×0a(x)w1−q

k (x, 0) dx− λ∫

Ω×0b(x)wp+1

k (x, 0) dx = 0 (4.6)


andf2k (w)‖wk + wφ‖2 − f1−q

k (w)∫

Ω×0a(x)(wk + wφ)1−q(x, 0)dx

− λfp+1k (w)

∫Ω×0

b(x)(wk + wφ)p+1(x, 0)dx = 0.(4.7)

Choose 0 < ρ < εk and w = ρv with ‖v‖ < 1. Then we find fk(w) such thatfk(0) = 1 and fk(w)(wk + wφ) ∈ N+

λ for all w ∈ Bρ(0).

Lemma 4.3. For λ ∈ (0,Λ) we have |〈f ′k(0), v〉| is finite for every 0 ≤ v ∈ Hs0,L(CΩ)

with ‖v‖ ≤ 1.

Proof. By (4.6) and (4.7) we have

0 = [f2k (w)− 1]‖wk + wφ‖2 + ‖wk + wφ‖2 − ‖wk‖2

− [f1−qk (w)− 1]

∫Ω×0

a(x)(wk + wφ)1−q(x, 0)dx

−∫

Ω×0a(x)[((wk + wφ)1−q − w1−q

k )(x, 0)]dx

− λ[fp+1k (w)− 1]

∫Ω×0

b(x)(wk + wφ)p+1(x, 0)dx

− λ∫

Ω×0b(x)[((wk + wφ)p+1 − wp+1

k )(x, 0)]dx

≤ [f2k (ρv)− 1]‖wk + ρvφ‖2 + ‖wk + ρvφ‖2 − ‖wk‖2

− [f1−qk (ρv)− 1]

∫Ω×0

a(x)(wk + ρvφ)1−q(x, 0)

− λ[fp+1k (ρv)− 1]

∫Ω×0

b(x)(wk + ρvφ)p+1(x, 0)

− λ∫

Ω×0b(x)[((wk + ρvφ)p+1 − wp+1

k )(x, 0)]

Dividing by ρ > 0 and passing to the limit ρ→ 0, we derive that

0 ≤ 2〈f ′k(0), v〉‖wk‖2 + 2ks∫CΩyc∇wk∇(vφ) dx dy

− (1− q)〈f ′k(0), v〉∫

Ω×0a(x)w1−q

k (x, 0)dx

− λ(p+ 1)〈f ′k(0), v〉∫

Ω×0b(x)wp+1

k (x, 0)dx

− λ(p+ 1)∫

Ω×0b(x)(wpkvφ)(x, 0)dx

= 〈f ′k(0), v〉[2‖wk‖2 − (1− q)

∫Ω×0

a(x)w1−qk (x, 0)dx

− λ(p+ 1)∫

Ω×0b(x)wp+1

k (x, 0)dx]

+ 2ks∫CΩyc∇wk∇(vφ) dx dy − λ(p+ 1)

∫Ω×0

b(x)(wpkvφ)(x, 0)dx


= 〈f ′k(0), v〉[(1 + q)‖wk‖2

− λ(p+ q)∫

Ω×0b(x)wp+1

k (x, 0)dx]

+ 2ks∫CΩyc∇wk∇(vφ) dx dy − λ(p+ 1)

∫Ω×0

b(x)(wpkvφ)(x, 0)dx. (4.8)

From this inequality and (4.4) we know that 〈f ′k(0), v〉 6= −∞. Now we show that〈f ′k(0), v〉 6= +∞. Arguing by contradiction, we assume that 〈f ′k(0), v〉 = +∞. Nowwe note that

|fk(ρv)− 1|‖wk‖+ fk(ρv)‖ρvφ‖ ≥ ‖[fk(ρv)− 1]wk + ρvfk(ρv)φ‖= ‖fk(ρv)(wk + ρvφ)− wk‖

(4.9)

and fk(ρv) > fk(0) = 1 for sufficiently large k.From the definition of derivative 〈f ′k(0), v〉, applying equation (4.2) with w =

fk(ρv)(wk + ρvφ) ∈ N+λ , we clearly have

[fk(ρv)− 1]‖wk‖k

+ fk(ρv)‖ρvφ‖k

≥ 1k‖fk(ρv)(wk + ρvφ)− wk‖

≥ Jλ(wk)− Jλ(fk(ρv)(wk + ρvφ))

=(1

2− 1

1− q

)‖wk‖2 + λ

( 11− q

− 1p+ 1

)∫Ω×0

b(x)wp+1k (x, 0)dx

+( 1

1− q− 1

2

)f2k (ρv)‖wk + ρvφ‖2

− λ(p+ q)(1− q)(p+ 1)

fp+1k (ρv)

∫Ω×0

b(x)(wk + ρvφ)p+1(x, 0)dx

=( 1

1− q− 1

2

)(‖wk + ρvφ‖2 − ‖wk‖2) +

( 11− q

− 12

)[f2k (ρv)− 1]‖wk + ρvφ‖2

− λ( 1

1− q− 1p+ 1

)fp+1k (ρv)

∫Ω×0

b(x)[((wk + ρvφ)p+1 − wp+1k )(x, 0)]dx

− λ( 1

1− q− 1p+ 1

)[fp+1k (ρv)− 1]

∫Ω×0

b(x)wp+1k (x, 0)dx.

Dividing by ρ > 0 and passing to the limit as ρ→ 0, we can obtain

〈f ′k(0), v〉‖wk‖k

+‖vφ‖k

≥(1 + q

1− q

)ks

∫CΩyc∇wk∇(vφ) dx dy +

(1 + q

1− q

)〈f ′k(0), v〉‖wk‖2

− λ(p+ q

1− q

)〈f ′k(0), v〉

∫Ω×0

b(x)wp+1k (x, 0)dx

− λ(p+ q

1− q

)∫Ω×0


=〈f ′k(0), v〉

1− q

[(1 + q)‖wk‖2 − λ(p+ q)

∫Ω×0

b(x)wp+1k (x, 0) dx

]


+(1 + q

1− q

)ks

∫CΩyc∇wk∇(vφ) dx dy − λ

(p+ q

1− q

)∫Ω×0

b(x)(wpkvφ)(x, 0)dx.

That is,

‖vφ‖k≥ 〈f

′k(0), v〉1− q

[(1 + q)‖wk‖2 − λ(p+ q)

∫Ω×0

b(x)wp+1k (x, 0) dx

− (1− q)‖wk‖k

]+(1 + q

1− q

)ks

∫CΩyc∇wk∇(vφ) dx dy

− λ(p+ q

1− q

)∫Ω×0

b(x)(wpkvφ)(x, 0)dx,

(4.10)

which is impossible because 〈f ′k(0), v〉 = +∞ and

(1+q)‖wk‖2−λ(p+q)∫

Ω×0b(x)wp+1

k (x, 0) dx− (1− q)‖wk‖k

≥ C2−(1− q)C1

k> 0.

In conclusion, |〈f ′k(0), v〉| < +∞. Furthermore, (4.4) with ‖wk‖ ≤ C1 and twoinequalities (4.8) and (4.10) also imply that

|〈f ′k(0), v〉| ≤ C3

for k sufficiently large and a suitable constant C3.

Lemma 4.4. For each 0 ≤ φ ∈ C∞0,L(CΩ) and for every 0 ≤ v ∈ Hs0,L(CΩ) with

‖v‖ ≤ 1, we have a(x)w−q0 vφ ∈ L1(Ω) and

ks

∫CΩyc∇w0∇(vφ) dx dy −

∫Ω×0

a(x)(w−q0 vφ)(x, 0)dx

− λ∫

Ω×0b(x)(wp0vφ)(x, 0)dx ≥ 0.

(4.11)

Proof. Applying (4.9) and (4.2) again, we obtain

[fk(ρv)− 1]‖wk‖k

+ fk(ρv)‖ρvφ‖k

≥ Jλ(wk)− Jλ(fk(ρv)(wk + ρvφ))

= −f2k (ρv)− 1

2‖wk‖2 −

f2k (ρv)

2(‖wk + ρvφ‖2 − ‖wk‖2)

+f1−qk (ρv)− 1

1− q

∫Ω×0

a(x)(wk + ρvφ)1−q(x, 0)

+1

1− q

∫Ω×0

a(x)[((wk + ρvφ)1−q − w1−qk )(x, 0)]

+ λfp+1k (ρv)− 1p+ 1

∫Ω×0

b(x)(wk + ρvφ)p+1(x, 0)

+λ

p+ 1

∫Ω×0

b(x)[((wk + ρvφ)p+1 − wp+1k )(x, 0)].

Dividing by ρ > 0 and passing to the limit ρ→ 0+, we obtain

|〈f ′k(0), v〉|‖wk‖k

+‖vφ‖k


≥ −〈f ′k(0), v〉‖wk‖2 − ks∫CΩyc∇wk∇(vφ) dx dy

+ 〈f ′k(0), v〉∫

Ω×0a(x)w1−q

k (x, 0)dx

+ lim infρ→0+

11− q

∫Ω×0

a(x)[((wk + ρvφ)1−q − w1−qk )(x, 0)]

ρdx

+ λ〈f ′k(0), v〉∫

Ω×0b(x)wp+1

k (x, 0)dx+ λ

∫Ω×0


= −〈f ′k(0), v〉[‖wk‖2 −

∫Ω×0

a(x)w1−qk (x, 0) dx− λ

∫Ω×0

b(x)wp+1k (x, 0) dx

]− ks

∫CΩyc∇wk∇(vφ) dx dy + λ

∫Ω×0

b(x)(wpkφ)(x, 0)dx

+ lim infρ→0+

11− q

∫Ω×0

a(x)[(wk + ρvφ)1−q − w1−qk )(x, 0)]

ρdx

= −ks∫CΩyc∇wk∇(vφ) dx dy + λ

∫Ω×0


+ lim infρ→0+

11− q

∫Ω×0

a(x)[((wk + ρvφ)1−q − w1−qk )(x, 0)]

ρdx,

Using above inequality, we have

lim infρ→0+

∫Ω×0

a(x)[((wk + ρvφ)1−q − w1−qk )(x, 0)]

ρdx

is finite. Now, since a(x)[((wk+vφ)1−q−w1−qk )(x, 0)] ≥ 0, then by Fatou’s Lemma,

we have∫Ω×0

a(x)(w−qk vφ)(x, 0)dx

≤ lim infρ→0+

11− q

∫Ω×0

a(x)[((wk + ρvφ)1−q − w1−qk )(x, 0)]

ρdx

≤ |〈f′k(0), v〉|‖wk‖+ ‖vφ‖

k+ ks

∫CΩyc∇wk∇(vφ) dx dy

− λ∫

Ω×0b(x)(wpkvφ)(x, 0)dx

≤ C1C3 + ‖vφ‖k

+ ks

∫CΩyc∇wk∇(vφ) dx dy − λ

∫Ω×0


Again using Fatou’s Lemma and this inequality, we have∫Ω×0

a(x)(w−q0 vφ)(x, 0)dx

≤∫

Ω×0

[lim infk→∞

a(x)(w−qk vφ)(x, 0)]dx

≤ lim infk→∞

∫Ω×0

a(x)(w−qk vφ)(x, 0)dx


≤ ks∫CΩyc∇w0∇(vφ) dx dy − λ

∫Ω×0

b(x)(wp0vφ)(x, 0)dx <∞,

which completes the proof.

Corollary 4.5. For every 0 ≤ φ ∈ Hs0,L(CΩ), we have a(x)w−q0 φ ∈ L1(Ω), w0 > 0

in CΩ and

ks

∫CΩyc∇w0∇φdx dy −

∫Ω×0

a(x)(w−q0 φ)(x, 0)dx

− λ∫

Ω×0b(x)(wp0φ)(x, 0)dx ≥ 0.

(4.12)

Proof. Choosing v ∈ Hs0,L(CΩ) such that v ≥ 0, v ≡ l in the neighborhood of

support of φ and ‖v‖ ≤ 1, for some l > 0 is a constant. Then by the Lemma 4.4,we note that

∫Ω×0 a(x)(w−q0 φ)(x, 0)dx < ∞, for every 0 ≤ φ ∈ C∞0,L(CΩ), which

guarantees that w0 > 0 a.e. in Ω×0. Also by the strong maximum principle [9],we obtain w0 > 0 in CΩ. Putting this choice of v in (4.11) we have

ks

∫CΩyc∇w0∇φdx dy − λ

∫Ω×0

b(x)(wp0φ)(x, 0)dx

−∫

Ω×0a(x)(w−q0 φ)(x, 0)dx ≥ 0.

for every 0 ≤ φ ∈ C∞0,L(CΩ). Hence by density argument, (4.12) holds for every0 ≤ φ ∈ Hs

0,L(CΩ), which completes the proof.

Lemma 4.6. We have w0 ∈ N+λ .

Proof. Using (4.12) with φ = w0, we obtain

‖w0‖2 ≥∫

Ω×0a(x)w1−q

0 (x, 0)dx+ λ

∫Ω×0

b(x)wp+10 (x, 0)dx.

On the other hand, by the weak lower semi-continuity of the norm, we have

‖w0‖2 ≤ lim infk→∞

‖wk‖2 ≤ lim supk→∞

‖wk‖2

=∫

Ω×0a(x)w1−q

0 (x, 0)dx+ λ

∫Ω×0

b(x)wp+10 (x, 0)dx.

Thus

‖w0‖2 =∫

Ω×0a(x)w1−q

0 (x, 0)dx+ λ

∫Ω×0

b(x)wp+10 (x, 0)dx. (4.13)

Consequently, wk → w0 in Hs0,L(CΩ) and w0 ∈ Nλ. Moreover, from (4.3) it follows

that

(1 + q)‖w0‖2 − λ(p+ q)∫

Ω×0b(x)wp+1

0 (x, 0)dx

=(1 + q)∫

Ω×0a(x)w1−q

0 (x, 0)dx− λ(p− 1)∫

Ω×0b(x)wp+1

0 (x, 0)dx > 0;

that is, w0 ∈ N+λ .

Lemma 4.7. The function w0 is a positive weak solution of problem (2.3).


Proof. Suppose that φ ∈ Hs0,L(CΩ) and ε > 0, we define Ψ = (w0 + εφ)+. Let

CΩ = Γ1 ∩ Γ2 with

Γ1 :=

(x, y) ∈ CΩ : (w0 + εφ)(x, y) > 0,Γ2 := (x, y) ∈ CΩ : (w0 + εφ)(x, y) ≤ 0

.

Let Ω× 0 = Ω1 × Ω2 with

Ω1 := (x, 0) ∈ Ω× 0 : (w0 + εφ)(x, 0) > 0,Ω2 := (x, 0) ∈ Ω× 0 : (w0 + εφ)(x, 0) ≤ 0.

Then Ω1 ⊂ Γ1, Ω2 ⊂ Γ2, Ψ|Γ1 = w0 + εφ, Ψ|Γ2 = 0, Ψ|Ω1(x, 0) = (w0 + εφ)(x, 0),and Ψ|Ω2(x, 0) = 0. Putting Ψ into (4.12) and using (4.13), we see that

0 ≤ ks∫CΩyc∇w0∇Ψ dx dy −

∫Ω×0

a(x)(w−q0 Ψ)(x, 0)dx

− λ∫

Ω×0b(x)(wp0Ψ)(x, 0)dx

= ks

∫Γ1

yc∇w0∇(w0 + εφ) dx dy −∫

Ω1

a(x)(w−q0 (w0 + εφ))(x, 0)dx

− λ∫

Ω1

b(x)(wp0(w0 + εφ))(x, 0)dx

= ks

∫CΩyc∇w0∇(w0 + εφ) dx dy −

∫Ω×0

a(x)(w−q0 (w0 + εφ)

)(x, 0)dx

− λ∫

Ω×0b(x) (wp0(w0 + εφ)) (x, 0)dx−

[ks

∫Γ2

yc∇w0∇(w0 + εφ) dx dy

−∫

Ω2

a(x)(w−q0 (w0 + εφ)

)(x, 0)dx− λ

∫Ω2

b(x) (wp0(w0 + εφ)) (x, 0)dx]

= ks

∫CΩyc|∇w0|2 dx dy −

∫Ω×0

a(x)w1−q0 (x, 0)dx− λ

∫Ω×0

b(x)wp+10 (x, 0)dx

+ ε[ks


∫Ω×0


− λ∫

Ω×0b(x)(wp0φ)(x, 0)dx

]− [ks

∫Γ2

yc|∇w0|2 −∫

Ω2

a(x)w1−q0 (x, 0)dx− λ

∫Ω2

b(x)wp+10 (x, 0)dx

]− ε[ks

∫Γ2

yc∇w0∇φdx dy −∫

Ω2


− λ∫

Ω2

b(x)(wp0φ)(x, 0)dx]

≤ ε[ks


∫Ω×0


− λ∫


]− εks

∫Γ2

yc∇w0∇φdx dy +∫

Ω2

a(x)w−q0 (w0 + εφ)dx


+ λ

∫Ω2

b(x)|εφ|p+1(x, 0)dx+ ελ

∫Ω2

b(x)(wp0φ)(x, 0)dx

≤ ε[ks


∫Ω×0


− λ∫


]− εks

∫Γ2

yc∇w0∇φdx dy + ελεp‖b‖L

2∗s2∗s−p−1 (Ω2)

(∫Ω2

|φ|2∗sdx) p+1

2∗s

+ ελ

∫Ω2

b(x)(wp0φ)(x, 0)dx.

Since the measure of Γ2 and Ω2 tend to zero as ε→ 0, it follows that∫Γ2

yc∇w0∇φdx dy → 0

as ε→ 0, and similarly for

λεp‖b‖L

2∗s2∗s−p−1 (Ω2)

(∫Ω2

|φ|2∗sdx) p+1

2∗s

and λ∫

Ω2b(x)(wp0φ)(x, 0)dx. Dividing by ε and letting ε→ 0, we obtain

ks


∫Ω×0


− λ∫

Ω×0b(x)(wp0φ)(x, 0)dx ≥ 0

and since this holds equally well for −φ, it follows that w0 is indeed a positive weaksolution of problem (2.3).

Lemma 4.8. There exists a minimizing sequence Wk in N−λ such that Wk →W0

strongly in N−λ . Moreover W0 is a positive weak solution of (2.3).

Proof. Using the Ekeland variational principle again, we may find a minimizingsequence Wk ⊂ N−λ for the minimizing problem infN−λ Jλ such that for Wk ∈Hs

0,L(CΩ), we have Wk W0 weakly in Hs0,L(CΩ) and pointwise a.e. in Ω × 0.

We can now repeat the argument used in Lemma 4.2 to derive that when λ ∈ (0,Λ)

(1+q)∫

Ω×0a(x)|W0(x, 0)|1−qdx−λ(p−1)

∫Ω×0

b(x)|W0(x, 0)|p+1dx < 0 (4.14)

which yields

(1 + q)∫

Ω×0a(x)|Wk(x, 0)|1−qdx− λ(p− 1)

∫Ω×0

b(x)|Wk(x, 0)|p+1dx ≤ −C4

for k sufficiently large and a suitable positive constant C4. At this point we mayproceed exactly as in Lemmas 4.3, 4.4, 4.6, 4.7 and Corollary 4.5, and conclude that0 < W0 ∈ Nλ is the required positive weak solution of problem (2.3). Moreoverfrom (4.14) it follows that

(1 + q)‖W0‖2 − λ(p+ q)∫

Ω×0b(x)W p+1

0 (x, 0)dx


= (1 + q)[ ∫

Ω×0a(x)W 1−q

0 (x, 0)dx+ λ

∫Ω×0

b(x)W p+10 (x, 0)dx

]− λ(p+ q)

∫Ω×0

b(x)W p+10 (x, 0)dx

= (1 + q)∫

Ω×0a(x)W 1−q

0 (x, 0)dx− λ(p− 1)∫

Ω×0b(x)W p+1

0 (x, 0)dx < 0,

that is W0 ∈ N−λ .

Proof of the Theorem 2.6. From Lemmas 4.7, 4.8 and 3.4, we can conclude thatproblem (2.3) has two positive weak solutions w0 ∈ N+

λ , W0 ∈ N−λ with ‖W0‖ >‖w0‖ for any λ ∈ (0,Λ). Hence, u0(·) = w0(·, 0) ∈ Hs

0(Ω) and U0(·) = W0(·, 0) ∈Hs

0(Ω) are positive solutions of the problem (1.1).

Proof of the Corollary 2.7. For any W ∈ N−λ , it follows from Lemma 3.4 that

‖W‖ > Aλ = Λ−1p−1

(1 + q

p+ q

)1/(p−1)( 1‖b‖

)1/(p−1)(√S) p+1p−1(Λλ

)1/(p−1)

.

Thus by the definition of Λ, and using 2(p+q)(1+q)(p−1) −

p+1p−1 = 1−q

1+q , we obtain,

‖W‖ >(

1 +1 + q

p− 1

)1/(1+q)

‖a‖1/(q+1)( 1√

S

) 1−q1+q(Λλ

)1/(p−1)

.

Hence, let Wε ∈ N−λ be the solution of problem (2.3) with p = 1 + ε, whereλ ∈ (0,Λ), we have

‖W‖ > Cε

(Λλ

)1/ε

,

where Cε =(1 + 1+q

ε

)1/(1+q)‖a‖1/(q+1)(

1√S

) 1−q1+q →∞ as ε→ 0. This completes the

proof of Corollary 2.7.

Acknowledgements. The first author is supported by the National Board forHigher Mathematics, Govt. of India, grant number: 2/40(2)/2015/R& D-II/5488.The second author is also supported by the National Board for Higher Mathematics,Govt. of India, 2/48(12)/2012/NBHM (R.P.)/R& D II/13095.

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Sarika Goyal

Department of Mathematics, Indian Institute of Technology Delhi, Hauz Khas, NewDelhi-16, India

E-mail address: [email protected]

Konijeti SreenadhDepartment of Mathematics, Indian Institute of Technology Delhi, Hauz Khas, New

Delhi-16, India

E-mail address: [email protected]

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FRACTIONAL ELLIPTIC EQUATIONS WITH SIGN-CHANGING …Key words and phrases. Non-local operator; singular nonlinearity; Nehari manifold; sign-changing weight function. c 2016 Texas State